Worked example 1: solution

Step 2 · Plan the route

One arrow links the units: cal → J. One conversion factor is needed.

Worked example 1: solution

The factor takes cal in the denominator, so cal cancels and J survives:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 2 · Plan the route

No single equality links kJ to cal. The route runs through the joule: kJ → J → cal. Two conversion factors are needed.

Worked example 2: solution

The prefix factor cancels kJ; the calorie factor takes J in the denominator:

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: C

D flipped the 4.184 factor: 1.10 × 10³ × 4.184 = 4.60 × 10³, and no unit cancels. A skipped the conversion: 1.10 × 10³ is still the count of joules. B also divided by 1000, as if the wanted unit were the food Calorie: 1.10 × 10³ / 4.184 / 1000 = 0.263.

Practice 2 · answer: B

A flipped the 4.184 factor: 6.20 × 10³ × 4.184 = 2.59 × 10⁴, and no unit cancels. C stopped after the prefix factor: 6.20 × 1000 = 6.20 × 10³, a count of joules, not calories. D treated kilojoules as joules: 6.20 / 4.184 = 1.48, a thousand times too small.

Worked example 3: solution

A common first attempt

Worked example 3: solution

A common first attempt

Worked example 3: the correct chain

Step 2 · Plan the route

The capital C marks the food Calorie: 1 Cal = 1000 cal. The route: Cal → cal → J → kJ. Three conversion factors are needed.

Worked example 3: the correct chain

Worked example 3: the correct chain

Worked example 3: the route on the map

Practice 3 · answer: B

A read Cal as cal: each value a thousand times too small. C flipped 4.184: 2.05 × 10⁵ / 4.184 = 4.90 × 10⁴ J. D wrote the kJ number as the J count: 858 kJ = 8.58 × 10⁵ J. E multiplied by 1000 at the last hop.

Practice 4 · answer: D

A read Cal as cal: 340. × 4.184 / 1000 = 1.42, a thousand times too small. B flipped the 4.184 factor: 340. × 1000 / 4.184 / 1000 = 81.3, and the joule never cancels. C stopped at Calories: 340. is the energy in Cal, never converted to kJ.

Worked example 1: solution

Step 1 · Name the system

The iron and oxygen are the system. The pouch, the air, your cold hands: surroundings.

Worked example 1: solution

ΔH is negative: −1648 kJ. Negative marks heat leaving the system.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the clue used

Worked example 2: solution

Step 1 · Name the system

The decomposing peroxide is the system; the cut, the skin, the air are surroundings.

Worked example 2: solution

Step 1 · Name the system
Step 2 · Find the heat's direction

The products sit 196 kJ below the reactants. That difference left the system as heat.

Worked example 2: solution

Step 1 · Name the system
Step 2 · Find the heat's direction
Step 3 · State the verdict

Worked example 2: solution

Step 1 · Name the system
Step 2 · Find the heat's direction
Step 3 · State the verdict

Worked example 2: the clue used

Your turn: the clue used

Practice 1 · answer: C

A flipped the sign convention: heat absorbed would be counted into the system, +198 kJ, not −198 kJ. B paired the right label with the wrong flow: exothermic means heat exits the system. D paired the right flow with the wrong label: a heat-releasing reaction is exothermic.

Worked example 3: solution

A common first answer

The cold skin is the evidence. Your hand is losing heat, and the hand is surroundings, not system.

Worked example 3: solution

The dissolving salt and water are the system. The pouch, your hand: surroundings.

Worked example 3: solution

Your hand cools: heat is leaving the surroundings and entering the system.

Worked example 3: solution

Worked example 3: solution

Practice 2: answer B

A took every change of state as heat in. Melting takes heat in, so its reverse, water turning to ice, gives heat out. C read evaporation's cooling as heat out. The cooling is the surroundings giving the evaporating water its heat. D assumed a reaction must absorb energy to happen; the acid and base mixture turns warm.

Practice 3: answer D

A called the higher level the bigger release: height is energy stored, so products above the reactants took energy in. B took the sign from the surroundings: the furnace loses 131 kJ, and the system gains it, +131 kJ. C paired the positive sign with the wrong label: heat in is endothermic.

Practice 4 · answer: A

B read a product-side heat term as heat gained: it leaves with the products, so it is released, −65 kJ, not +65 kJ. C matched the sign but not the label: a positive ΔH would mean heat in, and a bucket cannot turn hot from that. D matched the label but not the sign: endothermic and ΔH = −65 kJ contradict each other.

Extra practice 1 · answer: C

A took the freezing as the reaction's own change. Freezing does release heat, but the water is surroundings, and its heat went into the flask. B paired the label with the wrong flow: heat into the system is endothermic. D read the sign from the surroundings: the board loses heat, the system gains it, so ΔH is positive.

Extra practice 2 · answer: A

Three conversion factors are needed: kcal → cal → J → kJ.

B read the heat term from the wrong side: it sits with the reactants, so it is absorbed, +182 kJ. C flipped the 4.184 factor: 43.4 ÷ 4.184 = 10.4. D read kcal as cal: 43.4 × 4.184 = 182 J, only 0.182 kJ. E skipped the conversion: 43.4 is the heat in kcal, not kJ.

Worked example 1: solution

Step 1 · List the pieces

m = 250 g. c = 4.184 J/g·°C. ΔT = 50.0 − 20.0 = +30.0 °C. The unknown is q.

Worked example 1: solution

q already stands alone on its side of the equation; no rearranging is needed.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solving for q

Worked example 2: solution

Step 1 · List the pieces

q = 1347 J. m = 125 g. ΔT = 46.0 − 22.0 = +24.0 °C. The unknown is c.

Worked example 2: solution

Divide both sides by m·ΔT before any numbers go in.

Worked example 2: solution

No unit cancels here; they assemble into J/g·°C: the unit of a specific heat.

Worked example 2: solution

Worked example 2: identify the metal

Match the property

Only iron matches 0.449 J/g·°C. The block is iron.

Worked example 2: identify the metal

Step 4 · Check the sign

Worked example 2: solving for c

Practice 1 · answer: C

A left out the mass: 0.385 × 20.0 = 7.70 J warms one gram. B left out the temperature change: 75.0 × 0.385 = 28.9 J is one degree's worth. D divided by the specific heat: 75.0 × 20.0 ÷ 0.385 = 3.90 × 10³, with units g²·°C²/J.

Worked example 3: solution

A common first attempt

A lid that is releasing heat cannot end up hotter. The sign of q was dropped.

Worked example 3: solution

Released heat leaves the sample, so q = −8100 J. m = 150 g. c = 0.897 J/g·°C. The unknown is ΔT, then T_final.

Worked example 3: solution

Worked example 3: final temperature

Step 3 · Substitute and cancel units

Worked example 3: final temperature

Worked example 3: solving for ΔT, then T_final

Practice 2 · answer: B

A subtracted the temperatures in the wrong order: 62.0 − 22.0 = +40.0 °C gives +512 J, heat absorbed by a cooling sinker. C stopped at m × c: 100. × 0.128 = 12.8 J, one degree's worth with no sign. D left out the mass: 0.128 × (−40.0) = −5.12 J, the heat for a single gram.

Practice 3 · answer: C

A flipped the rearrangement: 45.0 × 4.184 ÷ 2260 = 0.0833 °C. B left out c: 2260 ÷ 45.0 = 50.2 °C. D multiplied by c: 2260 × 4.184 ÷ 45.0 = 210. °C. E left out the mass: 2260 ÷ 4.184 = 540. °C, the change for a single gram.

Practice 4 · answer: D

A left q in kilojoules: ΔT = 0.726 °C, so 27.7 °C. B subtracted T_initial: 725.6 − 27.0 = 699 °C; T_final = T_initial + ΔT. C stopped at ΔT: 726 °C is the change, not the final temperature. E multiplied by c: 2340 × 0.129 ÷ 25.0 = 12.07 °C, then 39.1 °C.

Practice 5 · answer: C

c is per gram, so the kilograms convert inside the chain: 2.0 kg is 2000 g. A used 2.0 as if it were grams: 2.0 × 0.89 × 157.0 = 280 J, a thousand times short. B left out the temperature change: 2000 × 0.89 = 1.8 × 10³ J, one degree's worth. D used 180.0 as ΔT: 2000 × 0.89 × 180.0 = 3.2 × 10⁵ J. ΔT is final minus initial: 157.0 °C.

Guided example: solution

Step 1 · Compare the specific heats

The masses are equal, so compare c alone. Gold's 0.129 is far smaller than water's 4.184.

Guided example: solution

Smaller c, larger ΔT. Gold swings more.

Guided example: solution

Guided example: solution

Guided example: the route on the strip

Practice 1 · answer: B

A read specific heat backwards: the larger c gives the smaller rise. C equal heat gives equal ΔT only when the specific heats match. D the heat sets how far each one warms, not which warms more: the smaller c wins at any heat.

Practice 2 · answer: D

A used aluminum's c: 450. / (30.0 × 0.897) = 16.7. B multiplied by c instead of dividing: 450. × 0.449 / 30.0 = 6.74. C dropped c: 450. / 30.0 = 15.0.

Practice 3 · answer: C

A ranked by c itself: the largest c gives the smallest ΔT, not the largest. B lumped the metals together: copper's 0.385 sits below iron's 0.449, so copper still swings further. D equal heat gives equal ΔT only when the specific heats match.

Worked example: solution

Step 1 · Compare the specific heats

Copper's c, 0.385, is smaller than aluminum's, 0.897. But mass is in the denominator too, and the copper sample is four times heavier.

Worked example: solution

Compare the whole sample's m·c: copper 200 × 0.385 = 77.0 J/°C, aluminum 50.0 × 0.897 = 44.85 J/°C. Aluminum's is smaller, so aluminum swings more.

Worked example: solution

Worked example: solution

Worked example: the route on the strip

Practice 4 · answer: A

B compared c alone: "smaller c wins" holds only at equal masses, and the lead block is heavier. C equal heat does not mean equal ΔT: the two samples' m·c differ. D computed both rises and pinned them on the wrong metals.

Extra practice 1 · answer: B

A took equal heat as an equal rise: 36.0 °C holds only when the specific heats match. C flipped the ratio: 36.0 × 0.385 / 0.128 = 108, but the larger c gives the smaller rise. D stopped halfway: 0.128 × 36.0 = 4.61 is the heat per gram, in J/g, not copper's rise.

Extra practice 2 · answer: A

B stopped in joules: 321 J is the right gap in the wrong unit. C put the factor upside down: 321 × 4.184 = 1.34 × 10³. D never subtracted silver's heat: 673.5 / 4.184 = 161 cal is iron's whole heat.

Extra practice 3 · answer: D

A took equal heat as an equal rise: that holds only when m·c matches, and 62.86 is not 35.88. B compared c alone: iron's 140. g makes its m·c the larger, so it rises only 23.9 °C. C stopped at the rise: 41.8 °C is how far aluminum climbs from 22.0 °C, not where it ends.

Guided example: solution

Step 1 · Set the balance

The water's numbers are complete, so its q comes first. The pipe's q is the same amount with the opposite sign.

Guided example: solution

Guided example: solution

Guided example: solution

Guided example: the route on the map

Worked example 1: solution

Step 1 · Set the balance

Each body keeps its own ΔT = T_f − T_i: 150 · (T_f − 70) = −100 · (T_f − 20). The hot side's ΔT will come out negative; the minus sign hands its released heat to the cold side.

Worked example 1: solution

The balance point is the mass-weighted average of the two starting temperatures. No distributing, no collecting terms: the masses weight the two starts directly.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Same signs on both heats is the slip. One body loses, the other gains.

Worked example 2: solution

Worked example 2: the final temperature

Step 2 · Solve without expanding

Worked example 2: the route on the map

Worked example 3: solution

Two moves are needed.

Step 1 · Heat gained by the water

The water gained 1.15 × 10⁵ J, so the burning food released it: q_food = −1.15 × 10⁵ J.

Worked example 3: solution

Worked example 3: the route on the map

Practice 1 · answer: D

A put the bearing's 45.0 g into the water's q: −(45.0 × 4.184 × 11.3) = −2.13 × 10³. B stopped at q_water: +5.20 × 10³ is the heat the water gained, before the sign flip. C used iron's specific heat for the water: −(110.0 × 0.449 × 11.3) = −558.

Practice 2 · answer: A

B counted one solution's 60.0 g: the whole 120.0 g mixture warmed, so −(60.0 × 4.184 × 13.1) = −3.29 × 10³ is half the heat. C stopped at q_solution: +6.58 × 10³, before the sign flip. D used the peak temperature as ΔT: −(120.0 × 4.184 × 33.9) = −1.70 × 10⁴.

Practice 2: the route on the map

Practice 3 · answer: A

B swapped the masses onto the wrong specific heats: 40.3 °C. C dropped the specific heats and weighted by mass alone: (60.0 × 120.0 + 125 × 25.0) ÷ 185 = 55.8 °C, a shortcut that works only when both bodies are water. D took the plain average of the starts: (120.0 + 25.0) ÷ 2 = 72.5 °C, as if the two bodies weighed the same and shared one c.

Practice 4 · answer: D

A dropped the minus in q_metal = −q_water: 2761 ÷ (95.0 × (−64.5)) = −0.451, a negative specific heat. B stopped at q_water: 2761 J, before dividing by the metal's m·ΔT. C used the final temperature as the metal's ΔT: 2761 ÷ (95.0 × 34.5) = 0.843. E swapped the two masses: −(95.0 × 4.184 × 16.5) ÷ (40.0 × (−64.5)) = 2.54.

Practice 5 · answer: A

B multiplied by 4.184: 1055 × 4.184 = 4.41 × 10³. C skipped the scale-up: 80.08 kJ ÷ 4.184 = 19.1, the chip alone. D used 32.1 °C as ΔT: 698. E stopped at kilojoules: 1.06 × 10³.

Worked example 1: solution

Step 1 · Match the target to the given

The target runs the other way: CO₂ breaks apart into C and O₂. It is the reverse.

Worked example 1: solution

Worked example 2: solution

Step 1 · Match the target to the given

The target is reversed and every coefficient is doubled. Two changes, so ΔH takes two steps.

Worked example 2: solution

Guided example: solution

Step 1 · Write ΔH per mole of C₃H₈

Propane's coefficient is 1: −2220 kJ per 1 mol C₃H₈. As a recipe, 0.250 mol is a quarter of a mole, so it releases a quarter of the heat.

Guided example: solution

Guided example: solution

Guided example: counted from the oxygen

Write ΔH per mole of O₂

Oxygen's coefficient is 5, so 5 mol O₂ goes under ΔH.

Guided example: counted from the oxygen

Guided example: counted from the oxygen

Guided example: the route on the map

Practice 1 · answer: C

A put HI's coefficient under ΔH: 2.50 × 53.0 ÷ 2 = +66.3 kJ, but I₂'s coefficient is 1. B gave the sign of a release: ΔH is positive, so heat is absorbed. D flipped the factor: 2.50 ÷ 53.0 = 0.0472, and mol never cancels.

Practice 2 · answer: B

D kept −92.2 kJ per 1 mol NH₃: 3.60 × −92.2 = −332 kJ, double the answer, because the equation forms 2 mol. A put H₂'s coefficient under ΔH: 3.60 × −92.2 ÷ 3 = −111 kJ. C flipped the factor: 3.60 × 2 ÷ −92.2 = −0.0781, and mol never cancels.

Worked example 3: solution

Two conversion factors are needed.

Step 1 · Pick the ΔH orientation

Two orientations exist. Only the one canceling mol CH₄ is used: −890 kJ / 1 mol CH₄, not 1 mol CH₄ / −890 kJ.

Worked example 3: solution

Worked example 3: the route on the map

Worked example 4: solution

Step 1 · Write ΔH per mole of SO₂

−198.2 kJ is the heat for the equation as written, which burns 2 mol of SO₂. Per mole of SO₂ that is −198.2/2 = −99.1 kJ. Keeping −198.2 per mole doubles the answer to −272 kJ: wrong.

Worked example 4: solution

Worked example 4: the route on the map

Practice 3 · answer: B

A scaled but kept the sign: 2 × 180 = +360 kJ. C flipped the sign but did not scale: −180 kJ. D left ΔH untouched at +180 kJ.

Practice 4 · answer: D

A kept −572 kJ per 1 mol H₂: 0.496 × −572 = −284 kJ, double the answer, because the equation burns 2 mol. B dropped the sign: burning releases heat, so ΔH is negative. C stopped at moles: 1.00 ÷ 2.016 = 0.496 mol, one factor short of kilojoules.

Practice 5 · answer: A

B kept 1203 kJ per 1 mol Mg: 2.50 × 10² ÷ 1203 × 24.31 = 5.05 g, half the true mass, because the equation burns 2 mol. C stopped at moles: 2.50 × 10² × 2/1203 = 0.416 mol Mg, one factor short of grams. D divided by the molar mass instead of multiplying: 0.416 ÷ 24.31 = 0.0171 g.

Practice 5: the route on the map

Practice 6 · answer: D

A flipped the ΔH factor: 0.306 mol × (2 mol / −10,941 kJ) = −5.60 × 10⁻⁵, and mol never cancels. B stopped at moles: 35.0 ÷ 114.22 = 0.306 mol, one factor short of kilojoules. C dropped the sign: combustion releases heat, so ΔH is negative. E used −10,941 kJ per 1 mol: 0.306 × (−10,941) = −3.35 × 10³ kJ, double the answer, because the equation burns 2 mol.